Step-by-Step Solution

1. Understand the Radioactive Decay Process

The problem describes the radioactive decay of Uranium-238 ($U$-238) into Lead-206 ($Pb$-206). Let's define the variables:

• $N_0$ = Initial number of moles (or atoms) of $U$-238.
• $N_t$ = Number of moles (or atoms) of $U$-238 remaining at time $t$.
• The number of moles of $U$-238 that have decayed is $(N_0 - N_t)$.

The decay process is $U\text{-}238 \rightarrow Pb\text{-}206$. The stoichiometry is 1:1. Therefore, the number of moles of $Pb$-206 formed at time $t$ is equal to the number of moles of $U$-238 that have decayed.

• Number of moles of $Pb$-206 at time $t$ = $(N_0 - N_t)$.

2. Relate Mass Ratio to Mole Ratio

We are given the ratio of the mass of $Pb$-206 to the mass of $U$-238 at time $t$.

• Mass of $U$-238 at time $t$: $m_U = N_t \times M_U$, where $M_U$ is the molar mass of $U$-238 (approximately 238 g/mol).
• Mass of $Pb$-206 at time $t$: $m_{Pb} = (N_0 - N_t) \times M_{Pb}$, where $M_{Pb}$ is the molar mass of $Pb$-206 (approximately 206 g/mol).

The given mass ratio is: $\frac{m_{Pb}}{m_U} = 7$ Substituting the expressions for the masses: $\frac{(N_0 - N_t) \times M_{Pb}}{N_t \times M_U} = 7$ $\frac{(N_0 - N_t) \times 206}{N_t \times 238} = 7$

3. Solve for the Ratio of Initial to Remaining Moles ($N_0/N_t$)

Rearrange the equation from Step 2 to solve for the ratio $N_0/N_t$. $\frac{N_0 - N_t}{N_t} = 7 \times \frac{238}{206}$ $\frac{N_0}{N_t} - 1 = \frac{1666}{206}$ $\frac{N_0}{N_t} = 1 + \frac{1666}{206} = \frac{206 + 1666}{206} = \frac{1872}{206}$ $\frac{N_0}{N_t} \approx 9.087$

4. Apply the Radioactive Decay Law

Radioactive decay is a first-order process. The age of the sample ($t$) can be calculated using the following formula, which relates the initial and remaining number of nuclei, the half-life ($t_{1/2}$), and the decay constant ($\\\lambda$): $N_t = N_0 e^{-\lambda t} \implies \ln\left(\frac{N_0}{N_t}\right) = \lambda t$ The decay constant is related to the half-life by $\\\lambda = \frac{\ln 2}{t_{1/2}}$. Substituting $\\\lambda$ into the decay equation: $t = \frac{1}{\lambda} \ln\left(\frac{N_0}{N_t}\right) = \frac{t_{1/2}}{\ln 2} \ln\left(\frac{N_0}{N_t}\right)$

5. Calculate the Age of the Sample

Substitute the known values into the equation for $t$:

• $t_{1/2} = 4.5 \times 10^9$ years
• $\\\ln 2 = 0.693$
• $\frac{N_0}{N_t} = \frac{1872}{206}$

$t = \frac{4.5 \times 10^9 \text{ years}}{0.693} \times \ln\left(\frac{1872}{206}\right)$ First, calculate the value of the natural logarithm: $\ln\left(\frac{1872}{206}\right) \approx \ln(9.087) \approx 2.2069$ Now, calculate the age $t$: $t \approx \frac{4.5 \times 10^9}{0.693} \times 2.2069$ $t \approx (6.4935 \times 10^9) \times 2.2069 \approx 14.33 \times 10^9 \text{ years}$

6. Determine the Value of P

The age of the sample is given in the format $\mathbf{P} \times 10^8$ years. We need to convert our calculated age to this format. $t = 14.33 \times 10^9 \text{ years} = 143.3 \times 10^8 \text{ years}$ Comparing this with $t = \mathbf{P} \times 10^8$ years, we get: $\mathbf{P} = 143.3$ Since the question asks for an integer value, we round the result to the nearest integer. $\mathbf{P} \approx 143$